# An object with a mass of 9 kg is on a plane with an incline of  - pi/4 . If it takes 15 N to start pushing the object down the plane and 5 N to keep pushing it, what are the coefficients of static and kinetic friction?

Nov 20, 2017

#### Answer:

$\setminus {\mu}_{s} = \left({F}_{a p p}^{s} / \left(m g \setminus \cos \setminus \theta\right) + \setminus \tan \setminus \theta\right) = 1.24$
$\setminus {\mu}_{k} = \left({F}_{a p p}^{k} / \left(m g \setminus \cos \setminus \theta\right) + \setminus \tan \setminus \theta\right) = 1.08$

#### Explanation:

Consider a body of mass $m$ on an incline with an inclination angle of $\setminus \theta$. To this a force is applied without disturbing the equilibrium. The forces acting on the object are
$\vec{W}$ : Weight of the object acting vertically downward;
${\vec{F}}_{a p p}$ : Force applied parallel to the inclined plane,
${\vec{F}}_{f}$ : Frictional force acting parallel to the plane of the incline,
$\vec{N}$ : Normal force acting perpendicular to the plane of the incline,

Consider a coordinate system with its X-axis parallel to the plane downward and Y-axis perpendicular to the plane upward.

Resolve the weight into components that are parallel and perpendicular to the inclined plane

vec W_{||} = -mg\sin\theta; \qquad vec W_{\_|_} = -mg\cos\theta

Equilibrium: $\setminus q \quad \vec{W} + {\vec{F}}_{a p p} + \vec{N} + {\vec{F}}_{f} = \vec{0}$

(A) Perpendicular Component: $\setminus q \quad \vec{N} + {\vec{W}}_{\setminus \bot} = \vec{0}$

N - W_{\_|_} = 0; \qquad N = mg\cos\theta;

(B) Parallel Component: $\setminus q \quad {\vec{W}}_{| |} + {F}_{a p p} + {\vec{F}}_{f} = \vec{0}$

W_{||} + F_{app} - F_f = 0; \qquad F_f = F_{app} + W_{||}  ...... (2)

Static Equilibrium: In this case the frictional force is due to static friction.

F_f = \mu_sN = \mu_s.mg\cos\theta; \qquad F_{app} = F_{app}^s = 15 N;

Expanding (2), $\setminus {\mu}_{s} . m g \setminus \cos \setminus \theta = {F}_{a p p}^{s} + m g \setminus \sin \setminus \theta$
$\setminus {\mu}_{s} = \left({F}_{a p p}^{s} / \left(m g \setminus \cos \setminus \theta\right) + \setminus \tan \setminus \theta\right)$ ...... (3)

Dynamic Equilibrium: In this case the frictional force is due to kinetic friction.

F_f = \mu_kN = \mu_k.mg\cos\theta; \qquad F_{app} = F_{app}^k = 5 N;

Expanding (2), $\setminus {\mu}_{k} . m g \setminus \cos \setminus \theta = {F}_{a p p}^{k} + m g \setminus \sin \setminus \theta$
$\setminus {\mu}_{k} = \left({F}_{a p p}^{k} / \left(m g \setminus \cos \setminus \theta\right) + \setminus \tan \setminus \theta\right)$ ...... (4)

Calculate:
\sin\theta = \cos\theta = 1/\sqrt{2}; \qquad \tan\theta = 1; \qquad mg\cos\theta = 62.37 $N$
mu_s = 1.24; \qquad \mu_k = 1.08